tions as you think necessary, by multiplying the square, of the mean diameter of each section by .7854.

Then to the sum of the areas of the two end sections, add four times the sum of the areas of all the even sections, and twice the sum of the areas of all the odd sections, not including the sections at the ends; multiply the sum by the common distance of the sections; divide the product by three; and the quotient will be the tent in cubic inches.

con

Divide the content thus found, by 282, 231, and 2150.42; and the respective quotients will be the content in ale and wine gallons, and malt bushels.

Note 1. Always make choice of an odd number of sections, in order that the number of parts into which the vessel is divided, may be equal. Seven or nine will, in general, be sufficient, except when the vessel is very deep, in which case it may be necessary to take eleven, thirteen, &c. as the case may require.

2. The perpendicular depth of the vessel must first be taken, in order to determine the common distance of the ordinates, which may be found by dividing the whole depth by the number of sections minus one.

3. Great care must be taken to obtain the diameters of the sections at equal perpendicular distances from each other; for if their distance be measured upon the side of the vessel, it is evident that the process will be incorrect. (See Problem III., Part VI., for the method of quartering circular vessels, taking cross diameters, &c. &c.)

4. The Rule given in this Problem is founded upon the method of equi-distant ordinates, described in Problem XX., Part IV. (See the Scholium to that Problem.)

RULE II.

Find the area corresponding to each mean diameter, in the Table of Ale or the Table of Wine Areas, Part VII.; and place those areas in regular succession, opposite to their respective diameters.

Then to the sum of the two extreme areas, add four times the sum of all the even areas, and twice the sum of all the odd areas; multiply this sum by the common distance of the diameters; divide the product by 3; and the quotient will be the content in ale or wine gallons. (See Problem XIII., Part IV.)

EXAMPLES.

1. What is the content in ale and wine gallons, of a circular vessel whose depth mn, is 120 inches; and the mean diameters of seven equi-distant, parallel sections as follow: viz., the diameter of the bottom or first section A B 124, the second CD 146, the third EF 161, the fourth GH 164, the fifth K L 166, the sixth MN 157, and the diameter, PR, of the top or seventh section 144 inches?

Here 1242 x.7854 15376x.7854 12076.3104, the area of the first section; 1462 x .7854-21316x.7854= 16741.5864, the area of the second section; 1619 x.7854 =25921 x.7854-20358.3534, the area of the third seetion; 1642 x.7854-26896 x .7854-21124.1184, the area of the fourth section; 1662 x .7854 27556x.7854 = 21642.4824, the area of the fifth section; 1572x.7854= 24649 x.7854-19359.3246, the area of the sixth section; and 1442 x.7854-20736×.7854-16286.0544, the area of the seventh or last section.

Then 12076.3104+16286.0544-28362.3648, the sum of the areas of the two end sections; (16741.5864+ 21124.1184+ 19359.3246) x 4 = 57225.0294 x 4 =

220

228900.1176, four times the sum of the areas of all the even sections; and (20358.3534+21642.4824) × 2=42000.8358 x2=84001.6716, twice the sum of the areas of all the odd

sections; also, =20, the common distance of the sec

120
6

tions; hence, (28362.3648+228900.1176+84001.6716)x 20 341264.154 x 20 6825283.08

3

3

tent in cubic inches; consequently

282

=2275094.36, the con

3

2275094.36

8067.710,

Then 42.8237+57.7518=100.5755, the sum of the two extreme areas; (59.3672 + 74.9080 + 68.6499) × 4 = 202.9251 × 4811.7004, four times the sum of all the even areas; and (72.1925 +76.7462) × 2 = 148.9387 × 2 = 297.8774, twice the sum of all the odd areas; hence 1201.1533 x 20

(100.5754+297.8774+811.7004) x

24203.0660 3

20

3

3

=8067.6886, the content in ale gallons, the

same as before.

Then 52.2784+70.5024=122.7808, the sum of the two extreme areas; (72.4744 + 91.4464 + 83.8066) × 4 = 247.7274×4=990.9096, four times the sum of all the even areas; and (88.1314+ 93.6904) × 2 = 181.8218 × 2 = 363.6436, twice the sum of all the odd areas; hence 20 1477.3340 x 20 (122.7808+990.9096+ 363.6436) x =

29546.6800

3

same as before.

3

3

=9848.8933, the content in wine gallons, the

2. The bottom or first diameter of a circular vessel is 93, the second 109.5, the third 120.8, the fourth 123.2, the fifth 124.5, the sixth 117.8, the seventh or top diameter 108.4, and the depth 114.6 inches; required the content in ale and wine gallons, by the second Rule.

Ans. 4340.13839 ale gallons, and 5298.37756 wine gallons.

3. The bottom or first diameter of a circular vessel measures 108.4, the second 123.6, the third 130.8, the fourth 136.6, the fifth 136.2, the sixth 131.8, the seventh 124.2, the eighth 114.8, the ninth or last 96.8, the depth 146.4, and the common distance of the diameters 18.3 inches; required the content in ale and wine gallons, and malt bushels, by the first Rule.

Ans. 6476.8996482 ale gallons, 7906.8645056 wine gallons, and 849.36231 malt bushels.

REMARK.

Sometimes the contents of vessels whose sides are curved, are found by dividing them into a number of

equal or unequal parts, and calculating the content of each part by the Rule for the frustum of a cone. This method, however, is extremely troublesome; and if the vessel be not divided into a great number of parts, so that the altitude of each part may be small, it is also very incorrect. For example, if the vessel belonging to the first Question in the last Problem, be taken as the frustums of six cones, its content will be 8034.0292 ale gallons, and 9807.776 wine gallons, which is too little by 33.6594 gallons of ale, and 41.1173 gallons of wine; hence we perceive the excellency of the Rules given in the last three Problems, for finding the contents of vessels when their sides are curved, and the nature of the curves cannot be determined; that is, when we cannot determine whether a vessel is the frustum of a sphere, the frustum of a spheroid, the frustum of a parabolic spindle, the frustum of a paraboloid, or the frustum of a hyperboloid.

PROBLEM XXII.

To find the content of the ungula or hoof of a cylinder, or the quantity of liquor contained in a cylindrical vessel placed in an inclining position, so that the surface of the liquor intersects the sides of the vessel in an oblique direction.

RULE.

By the Pen.

Multiply the square of the diameter of the base by half the sum of the greatest and least depths of the liquor; divide the product by 359.05 and 294.12, and the respective quotients will be the content in ale and wine gallons.

Note 1. The greatest and least depths of the liquor must not be taken perpendicularly to the horizon, but close to the sides of the vessel.

2. If a cylindrical vessel be placed in such an inclining position that the liquor just covers the bottom, it is evident that the hoof will be